Let A be a 3x3 matrix such that A^2 = A. If det(A) = 0 and the trace of A is 2, then the rank of A is:
Step-by-Step Solution
Key Concept: For an idempotent matrix A (A^2 = A), the eigenvalues are either 0 or 1. The trace of A is the sum of its eigenvalues, and the rank of A is the number of non-zero eigenvalues.
Since A^2 = A, the minimal polynomial divides x^2 - x = x(x-1), so the eigenvalues are 0 or 1. Let the eigenvalues be \lambda1, \lambda2, \lambda3. Given det(A) = 0, at least one eigenvalue is 0. Given trace(A) = \lambda1 + \lambda2 + \lambda3 = 2, and eigenvalues are 0 or 1, the eigenvalues must be 1, 1, 0. The rank of a diagonalizable matrix is the number of non-zero eigenvalues, which is 2.
Correct Answer: 2